`int(1)/(1+3 sin ^(2)x)dx` is equal to

Answer

To solve the integral \( I = \int \frac{1}{1 + 3 \sin^2 x} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{1}{1 + 3 \sin^2 x} \, dx \]

### Step 2: Divide by \(\cos^2 x\)
Next, we divide both the numerator and the denominator by \(\cos^2 x\):
\[ I = \int \frac{1/\cos^2 x}{\frac{1}{\cos^2 x} + 3 \frac{\sin^2 x}{\cos^2 x}} \, dx = \int \frac{\sec^2 x}{\sec^2 x + 3 \tan^2 x} \, dx \]

### Step 3: Substitute \( \tan x = t \)
Let \( t = \tan x \). Then, the derivative \( dt = \sec^2 x \, dx \) implies \( dx = \frac{dt}{\sec^2 x} \). Substituting these into the integral gives:
\[ I = \int \frac{dt}{1 + 3t^2} \]

### Step 4: Factor out the constant
We can factor out the constant from the denominator:
\[ I = \int \frac{dt}{1 + 3t^2} = \frac{1}{3} \int \frac{dt}{\frac{1}{3} + t^2} \]

### Step 5: Use the formula for integration
The integral \(\int \frac{dt}{a^2 + t^2} = \frac{1}{a} \tan^{-1} \left(\frac{t}{a}\right) + C\) can be applied here. Here, \( a = \frac{1}{\sqrt{3}} \):
\[ I = \frac{1}{3} \cdot \frac{1}{\frac{1}{\sqrt{3}}} \tan^{-1} \left(\frac{t}{\frac{1}{\sqrt{3}}}\right) + C = \frac{\sqrt{3}}{3} \tan^{-1} \left(\sqrt{3} t\right) + C \]

### Step 6: Substitute back \( t = \tan x \)
Now, substituting back \( t = \tan x \):
\[ I = \frac{\sqrt{3}}{3} \tan^{-1} \left(\sqrt{3} \tan x\right) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ I = \frac{\sqrt{3}}{3} \tan^{-1} \left(\sqrt{3} \tan x\right) + C \]

---